（1）由已知得：a_n+1=

an

3an+1

，

1

an+1

=

3an+1

an

=3+

1

an

∴

1

an+1

-

1

an

=3

∴{

1

an

}是首项为1，公差d=3的等差数列

（2）由（1）得

1

an

=1+（n-1）3=3n-2

∴S_n（x）=x+4x²+7x³+…+（3n-5）x^n-1+（3n-2）xⁿ

当x=1，S_n（1）=1+4+7+…+（3n-2）=

1+3n-2

2

•n=

n(3n-1)

2

当x≠1，0时，S_n（x）=x+4x²+7x³+…+（3n-5）x^n-1+（3n-2）xⁿ

xS_n（x）=x²+4x³+7x⁴+…+（3n-5）xⁿ+（3n-2）xⁿ⁺¹

（1-x）S_n（x）=x+（3x²+3x³+…+3xⁿ）-（3n-2）xⁿ⁺¹

=x+

3x2(1-xn-1)

1-x

-(3n-2)xn+1

∴Sn(x)=

x-(3n-2)xn+1

1-x

+

3x2(1-xn-1)

(1-x)2

=

x(1-x)-(3n-2)xn+1(1-x)+3x2(1-xn-1)

(1-x)2

=

(3n-2)xn+2-(3n-2)xn+1+x-x2+3x2-3xn+1

(1-x)2

=

(3n-2)xn+2-(3n+1)xn+1+2x2+x

(1-x)2

当x=0时，S_n（0）=0也适合．

综上所述，x=1，Sn(1)=

n(3n-1)

2

x≠1，S_n（x）=

(3n-2)xn+2-(3n+1)xn+1+2x2+x

(1-x)2

．